while doing some exercises about measurable cardinals, I got stuck on this one:
If $κ$ is the minimal cardinal that carries a non-trivial two-valued measure, then how can one prove that $κ$ is measurable?
I do not really have an idea on how to approach this, and am grateful for help.
Let $m:P(k)\to \{0,1\}$ be a non-trivial measure. Let $I=m^{-1}\{0\}.$
Suppose there exists cardinal $j$ with $\omega <j<k$ and $A=\{A_x:x<j\}\subset m^{-1}\{0\}$ such that $m(\cup A)=1.$
Let $i$ be the least such $j.$
Let $B=\{A_x\setminus \cup_{y<x}A_y: y\in i\}\setminus \{\emptyset\}.$ Then $m(\cup B)=1$ and $B\subset m^{-1}\{0\}.$ Now $B$ is a pair-wise disjoint family and $\emptyset \not \in B,$ and by the minimality of $j$ we have $|B|=j.$ So let $B=\{B_y:y<j\}.$
For $S\in P(j)$ let $m^*(S)=m(\cup_{y\in S}B_y).$ We may confirm that $m^*:P(j)\to \{0,1\}$ is a non-trivial measure. But $j<k$ so this contradicts the minimality of $k.$
So there is no such $j$.
Therefore $I=m^{-1}\{0\}$ is a free (non-principal) maximal ideal on $k,$ and $\cup A\in I$ whenever $A\subset P(k)$ and $|A|<k,$ so $F= \{k\setminus i:i\in I\}$ is a free maximal filter on $k,$ with $\cap C\in F$ whenever $C\subset F$ and $0<|C|<k.$
Remark: Regarding $m^*,$ if $S=\{S(n):n\in \omega\}\subset (m^*)^{-1}\{0\}$ then $S'=\{\cup_{y\in S(n)}B_y:n\in \omega\}\subset m^{-1}\{0\}, $ so $m^*(\cup S)=m(\cup S')=0.$